Copyright © 2009 Pearson Education, Inc. PowerPoint Lectures for Biology: Concepts & Connections, Sixth Edition Campbell, Reece, Taylor, Simon, and Dickey.

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Copyright © 2009 Pearson Education, Inc. PowerPoint Lectures for Biology: Concepts & Connections, Sixth Edition Campbell, Reece, Taylor, Simon, and Dickey Chapter 10 Molecular Biology of the Gene Lecture by Mary C. Colavito

10.2 DNA and RNA are polymers of nucleotides  The monomer unit of DNA and RNA is the nucleotide, containing –Nitrogenous base –5-carbon sugar –Phosphate group Copyright © 2009 Pearson Education, Inc.

 DNA and RNA are polymers called polynucleotides –A sugar-phosphate backbone is formed by covalent bonding between the phosphate of one nucleotide and the sugar of the next nucleotide –Nitrogenous bases extend from the sugar-phosphate backbone Copyright © 2009 Pearson Education, Inc. Animation: DNA and RNA Structure

Sugar-phosphate backbone DNA nucleotide Phosphate group Nitrogenous base Sugar DNA polynucleotide DNA nucleotide Sugar (deoxyribose) Thymine (T) Nitrogenous base (A, G, C, or T) Phosphate group

Sugar (deoxyribose) Thymine (T) Nitrogenous base (A, G, C, or T) Phosphate group

Pyrimidines Guanine (G) Adenine (A) Cytosine (C)Thymine (T) Purines

Sugar (ribose) Uracil (U) Nitrogenous base (A, G, C, or U) Phosphate group

Ribose Cytosine Uracil Phosphate Guanine Adenine

10.3 DNA is a double-stranded helix  James D. Watson and Francis Crick deduced the secondary structure of DNA, with X-ray crystallography data from Rosalind Franklin and Maurice Wilkins Copyright © 2009 Pearson Education, Inc.

 DNA is composed of two polynucleotide chains joined together by hydrogen bonding between bases, twisted into a helical shape –The sugar-phosphate backbone is on the outside –The nitrogenous bases are perpendicular to the backbone in the interior –Specific pairs of bases give the helix a uniform shape –A pairs with T, forming two hydrogen bonds –G pairs with C, forming three hydrogen bonds Copyright © 2009 Pearson Education, Inc. Animation: DNA Double Helix

Twist

Hydrogen bond Base pair Partial chemical structureComputer model Ribbon model

Base pair Ribbon model

Hydrogen bond Partial chemical structure

Computer model

DNA REPLICATION Copyright © 2009 Pearson Education, Inc.

10.4 DNA replication depends on specific base pairing  DNA replication follows a semiconservative model –The two DNA strands separate –Each strand is used as a pattern to produce a complementary strand, using specific base pairing –Each new DNA helix has one old strand with one new strand Copyright © 2009 Pearson Education, Inc. Animation: DNA Replication Overview

Parental molecule of DNA

Parental molecule of DNA Nucleotides Both parental strands serve as templates

Parental molecule of DNA Nucleotides Both parental strands serve as templates Two identical daughter molecules of DNA

 DNA replication begins at the origins of replication –DNA unwinds at the origin to produce a “bubble” –Replication proceeds in both directions from the origin –Replication ends when products from the bubbles merge with each other  DNA replication occurs in the 5’ 3’ direction –Replication is continuous on the 3’ 5’ template –Replication is discontinuous on the 5’ 3’ template, forming short segments 10.5 DNA replication proceeds in two directions at many sites simultaneously Copyright © 2009 Pearson Education, Inc.

Animation: Leading Strand 10.5 DNA replication proceeds in two directions at many sites simultaneously  Proteins involved in DNA replication –DNA polymerase adds nucleotides to a growing chain –DNA ligase joins small fragments into a continuous chain Copyright © 2009 Pearson Education, Inc. Animation: Lagging Strand Animation: DNA Replication Review Animation: Origins of Replication

Origin of replication Parental strand Daughter strand Bubble Two daughter DNA molecules

3 end 5 end 3 end 5 end  P P P P P P P P

Parental DNA 3 5 DNA polymerase molecule DNA ligase 3 5 Overall direction of replication Daughter strand synthesized continuously Daughter strand synthesized in pieces

THE FLOW OF GENETIC INFORMATION FROM DNA TO RNA TO PROTEIN Copyright © 2009 Pearson Education, Inc.

10.6 The DNA genotype is expressed as proteins, which provide the molecular basis for phenotypic traits  A gene is a sequence of DNA that directs the synthesis of a specific protein –DNA is transcribed into RNA –RNA is translated into protein  The presence and action of proteins determine the phenotype of an organism Copyright © 2009 Pearson Education, Inc.

10.6 The DNA genotype is expressed as proteins, which provide the molecular basis for phenotypic traits  Demonstrating the connections between genes and proteins –The one gene–one enzyme hypothesis was based on studies of inherited metabolic diseases –The one gene–one protein hypothesis expands the relationship to proteins other than enzymes –The one gene–one polypeptide hypothesis recognizes that some proteins are composed of multiple polypeptides Copyright © 2009 Pearson Education, Inc.

Cytoplasm Nucleus DNA

Cytoplasm Nucleus DNA Transcription RNA

Cytoplasm Nucleus DNA Transcription RNA Translation Protein

10.7 Genetic information written in codons is translated into amino acid sequences  The sequence of nucleotides in DNA provides a code for constructing a protein –Protein construction requires a conversion of a nucleotide sequence to an amino acid sequence –Transcription rewrites the DNA code into RNA, using the same nucleotide “language” –Each “word” is a codon, consisting of three nucleotides –Translation involves switching from the nucleotide “language” to amino acid “language” –Each amino acid is specified by a codon –64 codons are possible –Some amino acids have more than one possible codon Copyright © 2009 Pearson Education, Inc.

Polypeptide Translation Transcription Gene 1 DNA molecule DNA strand Codon Amino acid Gene 2 Gene 3 RNA

Polypeptide Translation Transcription DNA strand Codon Amino acid RNA

10.8 The genetic code is the Rosetta stone of life  Characteristics of the genetic code –Triplet: Three nucleotides specify one amino acid –61 codons correspond to amino acids –AUG codes for methionine and signals the start of transcription –3 “stop” codons signal the end of translation Copyright © 2009 Pearson Education, Inc.

10.8 The genetic code is the Rosetta stone of life –Redundant: More than one codon for some amino acids –Unambiguous: Any codon for one amino acid does not code for any other amino acid –Does not contain spacers or punctuation: Codons are adjacent to each other with no gaps in between –Nearly universal Copyright © 2009 Pearson Education, Inc.

First base Third base Second base

Strand to be transcribed DNA

Strand to be transcribed DNA Start codon RNA Transcription Stop codon

Strand to be transcribed DNA Start codon RNA Transcription Stop codon Polypeptide Translation Met Lys Phe

10.9 Transcription produces genetic messages in the form of RNA  Overview of transcription –The two DNA strands separate –One strand is used as a pattern to produce an RNA chain, using specific base pairing –For A in DNA, U is placed in RNA –RNA polymerase catalyzes the reaction Copyright © 2009 Pearson Education, Inc.

10.9 Transcription produces genetic messages in the form of RNA  Stages of transcription –Initiation: RNA polymerase binds to a promoter, where the helix unwinds and transcription starts –Elongation: RNA nucleotides are added to the chain –Termination: RNA polymerase reaches a terminator sequence and detaches from the template Copyright © 2009 Pearson Education, Inc. Animation: Transcription

RNA polymerase Newly made RNA Direction of transcription Template strand of DNA RNA nucleotides

Terminator DNA DNA of gene RNA polymerase Initiation Promoter DNA 1 Elongation 2 Area shown in Figure 10.9A Termination 3 Growing RNA polymerase Completed RNA

10.10 Eukaryotic RNA is processed before leaving the nucleus  Messenger RNA (mRNA) contains codons for protein sequences  Eukaryotic mRNA has interrupting sequences called introns, separating the coding regions called exons  Eukaryotic mRNA undergoes processing before leaving the nucleus –Cap added to 5’ end: single guanine nucleotide –Tail added to 3’ end: Poly-A tail of 50–250 adenines –RNA splicing: removal of introns and joining of exons to produce a continuous coding sequence Copyright © 2009 Pearson Education, Inc.

RNA transcript with cap and tail Exons spliced together Introns removed Transcription Addition of cap and tail Tail DNA mRNA Cap Exon Intron Coding sequence Nucleus Cytoplasm

10.11 Transfer RNA molecules serve as interpreters during translation  Transfer RNA (tRNA) molecules match an amino acid to its corresponding mRNA codon –tRNA structure allows it to convert one language to the other –An amino acid attachment site allows each tRNA to carry a specific amino acid –An anticodon allows the tRNA to bind to a specific mRNA codon, complementary in sequence –A pairs with U, G pairs with C Copyright © 2009 Pearson Education, Inc.

Anticodon Amino acid attachment site RNA polynucleotide chain Hydrogen bond

10.12 Ribosomes build polypeptides  Translation occurs on the surface of the ribosome –Ribosomes have two subunits: small and large –Each subunit is composed of ribosomal RNAs and proteins –Ribosomal subunits come together during translation –Ribosomes have binding sites for mRNA and tRNAs Copyright © 2009 Pearson Education, Inc.

tRNA molecules Growing polypeptide Large subunit Small subunit mRNA

tRNA-binding sites Large subunit Small subunit mRNA binding site

mRNA Next amino acid to be added to polypeptide Growing polypeptide Codons tRNA

10.13 An initiation codon marks the start of an mRNA message  Initiation brings together the components needed to begin RNA synthesis  Initiation occurs in two steps 1.mRNA binds to a small ribosomal subunit, and the first tRNA binds to mRNA at the start codon –The start codon reads AUG and codes for methionine –The first tRNA has the anticodon UAC 2.A large ribosomal subunit joins the small subunit, allowing the ribosome to function –The first tRNA occupies the P site, which will hold the growing peptide chain –The A site is available to receive the next tRNA Copyright © 2009 Pearson Education, Inc.

Start of genetic message End

Small ribosomal subunit Start codon P site mRNA A site Large ribosomal subunit Initiator tRNA Met 2 1

10.14 Elongation adds amino acids to the polypeptide chain until a stop codon terminates translation  Elongation is the addition of amino acids to the polypeptide chain  Each cycle of elongation has three steps 1.Codon recognition: next tRNA binds to the mRNA at the A site 2.Peptide bond formation: joining of the new amino acid to the chain –Amino acids on the tRNA at the P site are attached by a covalent bond to the amino acid on the tRNA at the A site Copyright © 2009 Pearson Education, Inc.

3.Translocation: tRNA is released from the P site and the ribosome moves tRNA from the A site into the P site Copyright © 2009 Pearson Education, Inc Elongation adds amino acids to the polypeptide chain until a stop codon terminates translation

 Elongation continues until the ribosome reaches a stop codon  Applying Your Knowledge How many cycles of elongation are required to produce a protein with 100 amino acids?  Termination –The completed polypeptide is released –The ribosomal subunits separate –mRNA is released and can be translated again Copyright © 2009 Pearson Education, Inc Elongation adds amino acids to the polypeptide chain until a stop codon terminates translation Animation: Translation

Polypeptide A site 1 Codon recognition Codons Amino acid Anticodon P site mRNA

Polypeptide A site 1 Codon recognition Codons Amino acid Anticodon P site mRNA 2 Peptide bond formation

Polypeptide A site 1 Codon recognition Codons Amino acid Anticodon P site mRNA 2 Peptide bond formation 3 Translocation New peptide bond

Polypeptide A site 1 Codon recognition Codons Amino acid Anticodon P site mRNA 2 Peptide bond formation 3 Translocation New peptide bond Stop codon mRNA movement

10.15 Review: The flow of genetic information in the cell is DNA  RNA  protein  Does translation represent: –DNA  RNA or RNA  protein?  Where does the information for producing a protein originate: –DNA or RNA?  Which one has a linear sequence of codons: –rRNA, mRNA, or tRNA?  Which one directly influences the phenotype: –DNA, RNA, or protein? Copyright © 2009 Pearson Education, Inc.

Each amino acid attaches to its proper tRNA with the help of a specific enzyme and ATP. mRNA is transcribed from a DNA template. 2 1 RNA polymerase Amino acid DNA Transcription mRNA tRNA ATP Translation Enzyme 3 The mRNA, the first tRNA, and the ribo- somal sub-units come together. Initiator tRNA Large ribosomal subunit Anticodon Initiation of polypeptide synthesis Small ribosomal subunit mRNA Start Codon New peptide bond forming Growing polypeptide 4 A succession of tRNAs add their amino acids to the polypeptide chain as the mRNA is moved through the ribosome, one codon at a time. Elongation Codons mRNA Polypeptide 5 The ribosome recognizes a stop codon. The poly- peptide is terminated and released. Termination Stop codon

mRNA is transcribed from a DNA template. RNA polymerase Each amino acid attaches to its proper tRNA with the help of a specific enzyme and ATP. Amino acid DNA Transcription mRNA tRNA ATP Translation Enzyme The mRNA, the first tRNA, and the ribosomal sub-units come together. Initiator tRNA Large ribosomal subunit Anticodon Initiation of polypeptide synthesis Small ribosomal subunit mRNA Start Codon 1 2 3

New peptide bond forming Growing polypeptide 4 A succession of tRNAs add their amino acids to the polypeptide chain as the mRNA is moved through the ribosome, one codon at a time. Elongation Codons mRNA Polypeptide 5 The ribosome recognizes a stop codon. The polypeptide is terminated and released. Termination Stop codon

10.16 Mutations can change the meaning of genes  A mutation is a change in the nucleotide sequence of DNA –Base substitutions: replacement of one nucleotide with another –Effect depends on whether there is an amino acid change that alters the function of the protein –Deletions or insertions –Alter the reading frame of the mRNA, so that nucleotides are grouped into different codons –Lead to significant changes in amino acid sequence downstream of mutation –Cause a nonfunctional polypeptide to be produced Copyright © 2009 Pearson Education, Inc.

10.16 Mutations can change the meaning of genes  Mutations can be –Spontaneous: due to errors in DNA replication or recombination –Induced by mutagens –High-energy radiation –Chemicals Copyright © 2009 Pearson Education, Inc.

Normal hemoglobin DNAMutant hemoglobin DNA Sickle-cell hemoglobin Normal hemoglobin mRNA ValGlu

Normal gene Protein Base substitution Base deletion Missing mRNA Met Lys Phe Ser Ala Met Lys Phe Gly Ala Met Lys Leu Ala His

MICROBIAL GENETICS Copyright © 2009 Pearson Education, Inc.

10.17 Viral DNA may become part of the host chromosome  Viruses have two types of reproductive cycles –Lytic cycle –Viral particles are produced using host cell components –The host cell lyses, and viruses are released Copyright © 2009 Pearson Education, Inc.

10.17 Viral DNA may become part of the host chromosome  Viruses have two types of reproductive cycles –Lysogenic cycle –Viral DNA is inserted into the host chromosome by recombination –Viral DNA is duplicated along with the host chromosome during each cell division –The inserted phage DNA is called a prophage –Most prophage genes are inactive –Environmental signals can cause a switch to the lytic cycle Copyright © 2009 Pearson Education, Inc. Animation: Phage T4 Lytic Cycle Animation: Phage Lambda Lysogenic and Lytic Cycles

Bacterial chromosome Phage injects DNA Phage Phage DNA Attaches to cell Phage DNA circularizes Lytic cycle 4 New phage DNA and proteins are synthesized Phages assemble Cell lyses, releasing phages

Bacterial chromosome Phage injects DNA Phage Phage DNA Attaches to cell Phage DNA circularizes Lytic cycle 4 New phage DNA and proteins are synthesized Phages assemble Cell lyses, releasing phages Phage DNA inserts into the bacterial chromosome by recombination Lysogenic bacterium reproduces normally, replicating the prophage at each cell division Prophage Lysogenic cycle Many cell divisions OR

Bacterial chromosome Phage injects DNA Phage Phage DNA Attaches to cell Phage DNA circularizes Lytic cycle New phage DNA and proteins are synthesized Phages assemble Cell lyses, releasing phages

Bacterial chromosome Phage injects DNA Phage DNA circularizes Phage DNA inserts into the bacterial chromosome by recombination Lysogenic bacterium reproduces normally, replicating the prophage at each cell division Prophage Lysogenic cycle Many cell divisions Phage Phage DNA Attaches to cell 1

10.18 CONNECTION: Many viruses cause disease in animals and plants  Both DNA viruses and RNA viruses cause disease in animals  Reproductive cycle of an RNA virus –Entry –Glycoprotein spikes contact host cell receptors –Viral envelope fuses with host plasma membrane –Uncoating of viral particle to release the RNA genome –mRNA synthesis using a viral enzyme –Protein synthesis –RNA synthesis of new viral genome –Assembly of viral particles Copyright © 2009 Pearson Education, Inc.

10.18 CONNECTION: Many viruses cause disease in animals and plants  Some animal viruses reproduce in the cell nucleus  Most plant viruses are RNA viruses –They breach the outer protective layer of the plant –They spread from cell to cell through plasmodesmata –Infection can spread to other plants by animals, humans, or farming practices Copyright © 2009 Pearson Education, Inc. Animation: Simplified Viral Reproductive Cycle

Plasma membrane of host cell VIRUS Entry Uncoating Viral RNA (genome) Viral RNA (genome) Membranous envelope Protein coat Glycoprotein spike RNA synthesis by viral enzyme Template RNA synthesis (other strand) Protein synthesis mRNA New viral genome New viral proteins Assembly 7 Exit

Plasma membrane of host cell VIRUS Entry Viral RNA (genome) Viral RNA (genome) 2 Membranous envelope Protein coat Glycoprotein spike Uncoating RNA synthesis by viral enzyme 3 1

Template RNA synthesis (other strand) Protein synthesis New viral genome mRNA New viral proteins Assembly Exit

10.19 EVOLUTION CONNECTION: Emerging viruses threaten human health  How do emerging viruses cause human diseases? –Mutation –RNA viruses mutate rapidly –Contact between species –Viruses from other animals spread to humans –Spread from isolated populations Copyright © 2009 Pearson Education, Inc.

10.19 EVOLUTION CONNECTION: Emerging viruses threaten human health  Examples of emerging viruses –HIV –Ebola virus –West Nile virus –RNA coronavirus causing severe acute respiratory syndrome (SARS) –Avian flu virus Copyright © 2009 Pearson Education, Inc.

10.20 The AIDS virus makes DNA on an RNA template  AIDS is caused by HIV, human immunodeficiency virus  HIV is a retrovirus, containing –Two copies of its RNA genome –Reverse transcriptase, an enzyme that produces DNA from an RNA template Copyright © 2009 Pearson Education, Inc.

 HIV duplication –Reverse transcriptase uses RNA to produce one DNA strand –Reverse transcriptase produces the complementary DNA strand –Viral DNA enters the nucleus and integrates into the chromosome, becoming a provirus –Provirus DNA is used to produce mRNA –mRNA is translated to produce viral proteins –Viral particles are assembled and leave the host cell Copyright © 2009 Pearson Education, Inc The AIDS virus makes DNA on an RNA template Animation: HIV Reproductive Cycle

Reverse transcriptase RNA (two identical strands) Protein coat Glycoprotein Envelope

Double- stranded DNA Viral RNA and proteins DNA strand Viral RNA N UCLEUS C YTOPLASM Chromosomal DNA Provirus DNA RNA

10.21 Viroids and prions are formidable pathogens in plants and animals  Some infectious agents are made only of RNA or protein –Viroids: circular RNA molecules that infect plants –Replicate within host cells without producing proteins –Interfere with plant growth –Prions: infectious proteins that cause brain diseases in animals –Misfolded forms of normal brain proteins –Convert normal protein to misfolded form Copyright © 2009 Pearson Education, Inc.

10.22 Bacteria can transfer DNA in three ways  Three mechanisms allow transfer of bacterial DNA –Transformation is the uptake of DNA from the surrounding environment –Transduction is gene transfer through bacteriophages –Conjugation is the transfer of DNA from a donor to a recipient bacterial cell through a cytoplasmic bridge  Recombination of the transferred DNA with the host bacterial chromosome leads to new combinations of genes Copyright © 2009 Pearson Education, Inc.

DNA enters cell Bacterial chromosome (DNA) Fragment of DNA from another bacterial cell

Phage Fragment of DNA from another bacterial cell (former phage host)

Mating bridge Sex pili Donor cell (“male”) Recipient cell (“female”)

Donated DNA Recipient cell’s chromosome Crossovers Recombinant chromosome Degraded DNA

10.23 Bacterial plasmids can serve as carriers for gene transfer  Plasmids are small circular DNA molecules that are separate from the bacterial chromosome –F factor is involved in conjugation –When integrated into the chromosome, transfers bacterial genes from donor to recipient –When separate, transfers F-factor plasmid –R plasmids transfer genes for antibiotic resistance by conjugation Copyright © 2009 Pearson Education, Inc.

Male (donor) cell Origin of F replication Bacterial chromosome F factor starts replication and transfer of chromosome F factor (integrated) Recipient cell Only part of the chromosome transfers Recombination can occur

Male (donor) cell Bacterial chromosome F factor starts replication and transfer F factor (plasmid) Plasmid completes transfer and circularizes Cell now male

Plasmids

Sugar- phosphate backbone Deoxy- ribose Ribose Nucleotide Sugar Phosphate group DNA Nitrogenous base Nitrogenous base Polynucleotide DNA RNA Sugar CGATCGAT CGAUCGAU

Codons Growing polypeptide Amino acid tRNA Anticodon Large ribosomal subunit mRNA Small ribosomal subunit

 Viruses are invaders that sabotage our cells –Viruses have genetic material surrounded by a protein coat and, in some cases, a membranous envelope –Viral proteins bind to receptors on a host’s target cell –Viral nucleic acid enters the cell –It may remain dormant by integrating into a host chromosome –When activated, viral DNA triggers viral duplication, using the host’s molecules and organelles –The host cell is destroyed, and newly replicated viruses are released to continue the infection Introduction: Sabotage Inside Our Cells Copyright © 2009 Pearson Education, Inc.

THE STRUCTURE OF THE GENETIC MATERIAL Copyright © 2009 Pearson Education, Inc.

10.1 Experiments showed that DNA is the genetic material  Frederick Griffith discovered that a “transforming factor” could be transferred into a bacterial cell –Disease-causing bacteria were killed by heat –Harmless bacteria were incubated with heat-killed bacteria –Some harmless cells were converted to disease- causing bacteria, a process called transformation –The disease-causing characteristic was inherited by descendants of the transformed cells Copyright © 2009 Pearson Education, Inc.

10.1 Experiments showed that DNA is the genetic material  Alfred Hershey and Martha Chase used bacteriophages to show that DNA is the genetic material –Bacteriophages are viruses that infect bacterial cells –Phages were labeled with radioactive sulfur to detect proteins or radioactive phosphorus to detect DNA –Bacteria were infected with either type of labeled phage to determine which substance was injected into cells and which remained outside Copyright © 2009 Pearson Education, Inc.

10.1 Experiments showed that DNA is the genetic material –The sulfur-labeled protein stayed with the phages outside the bacterial cell, while the phosphorus-labeled DNA was detected inside cells –Cells with phosphorus-labeled DNA produced new bacteriophages with radioactivity in DNA but not in protein Copyright © 2009 Pearson Education, Inc. Animation: Hershey-Chase Experiment Animation: Phage T2 Reproductive Cycle

Head Tail fiber DNA Tail

Head Tail fiber DNA Tail

Batch 1 Radioactive protein Bacterium Radioactive protein DNA Phage Pellet Radioactive DNA Batch 2 Radioactive DNA Empty protein shell Phage DNA Centrifuge Radioactivity in liquid Measure the radioactivity in the pellet and the liquid. 4 Centrifuge the mixture so bacteria form a pellet at the bottom of the test tube. 3 Agitate in a blender to separate phages outside the bacteria from the cells and their contents. 2 Mix radioactively labeled phages with bacteria. The phages infect the bacterial cells. 1 Pellet Centrifuge Radioactivity in pellet

Batch 1 Radioactive protein Bacterium Radioactive protein DNA Phage Radioactive DNA Batch 2 Radioactive DNA Agitate in a blender to separate phages outside the bacteria from the cells and their contents. 2 Mix radioactively labeled phages with bacteria. The phages infect the bacterial cells. 1 Empty protein shell Phage DNA

Pellet Empty protein shell Phage DNA Centrifuge Radioactivity in liquid Measure the radioactivity in the pellet and the liquid. Centrifuge the mixture so bacteria form a pellet at the bottom of the test tube. Pellet Centrifuge Radioactivity in pellet 4 3

Phage attaches to bacterial cell. Phage injects DNA.Phage DNA directs host cell to make more phage DNA and protein parts. New phages assemble. Cell lyses and releases new phages.

comes in three kinds called RNA (d) (e) (f) is performed by organelles called use amino-acid-bearing molecules called (h) molecules are components of one or more polymers made from monomers called is performed by enzyme called is a polymer made from monomers called DNA (a) (b) (c) Protein (g) (i)

1.Compare and contrast the structures of DNA and RNA 2.Describe how DNA replicates 3.Explain how a protein is produced 4.Distinguish between the functions of mRNA, tRNA, and rRNA in translation 5.Determine DNA, RNA, and protein sequences when given any complementary sequence You should now be able to Copyright © 2009 Pearson Education, Inc.

6.Distinguish between exons and introns and describe the steps in RNA processing that lead to a mature mRNA 7.Explain the relationship between DNA genotype and the action of proteins in influencing phenotype 8.Distinguish between the effects of base substitution and insertion or deletion mutations You should now be able to Copyright © 2009 Pearson Education, Inc.

9.Distinguish between lytic and lysogenic viral reproductive cycles and describe how RNA viruses are duplicated within a host cell 10.Explain how an emerging virus can become a threat to human health 11.Identify three methods of transfer for bacterial genes 12.Distinguish between viroids and prions 13.Describe the effects of transferring plasmids from donor to recipient cells You should now be able to Copyright © 2009 Pearson Education, Inc.